Prove isomorphism between two representations $G$: a $l$-group. $(\pi,V),(\sigma,W)$:  admissible repns of $G$. $\exists$ nondegenarate bilinear form $\langle \cdot,\cdot\rangle:V\times W\to\mathbb{C}$ which is $G$-invariant, i.e. $\langle \pi_g(v),\sigma_g(w)\rangle=\langle v,w\rangle$, $\forall g\in G, (v,w)\in V\times W$. Show that $\sigma\cong \pi^{\vee}$, where $\pi^{\vee}$ is the contragradient repn of $\pi$.
 A: Hint: In linear algebra, one learns about a natural map
$$\text{bilinear forms on } V \times W \rightarrow Hom(V, W^*),$$
where $V, W$ are arbitrary (maybe finite-dimensional) vector spaces, and $W^*$ is the dual of $W$. This map is an iso in interesting cases.
Now think about what Hom's we get on the right if we only look at $G$-invariant forms on the left; and then think about what non-degeneracy on the left corresponds to on the right.
A: Let $\widetilde{V}$ be the underlying space of the contragredient $\tilde{\pi}$ of $\pi$. There is an obvious $G$-linear map
$$\Phi: W \rightarrow \widetilde{V}$$
defined by $\Phi(w)(v) = \langle v,w \rangle$.  The problem is to show that this is a bijection.  Use the fact that a sequence
$$0 \rightarrow W' \rightarrow W \rightarrow W'' \rightarrow 0$$
of representations of $G$ is exact if and only if
$$0 \rightarrow W'^K \rightarrow W^K \rightarrow W''^K \rightarrow 0$$
is exact for all open compact subgroups $K$ of $G$.  In this way you can reduce to the finite dimensional case.
Edit: as indicated in the comments, one nontrivial thing that needs to be checked is that for every open compact subgroup $K$ of $G$, the restriction of the pairing to $V^K \times W^K$ remains nondegenerate.  In other words, we need to check that if $v \in V^K$, and $\langle v, w \rangle = 0$ for all $w \in W^K$, then $v = 0$.
This follows from the fact that, since $K$ is compact, the restriction of $(\sigma, W)$ to $K$ decomposes into a direct sum of irreducible representations.  Let $L$ be one of these irreducible representations of $K$.  We are done if we can show that $\langle v, w \rangle = 0$ whenever $v \in V^K$ and $w \in L$.  
If we fix $v \in V^K$, it is clear that $w \mapsto \langle v,w \rangle$ is a $K$-invariant linear functional $L \rightarrow \mathbb C$.  The kernel of this functional must be equal to $L$, since $L$ is irreducible. This implies that $\langle v, w \rangle = 0$ for all $w \in L$.  
